Problem: Is ${964523}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {964523}= &&{9}\cdot100000+ \\&&{6}\cdot10000+ \\&&{4}\cdot1000+ \\&&{5}\cdot100+ \\&&{2}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {964523}= &&{9}(99999+1)+ \\&&{6}(9999+1)+ \\&&{4}(999+1)+ \\&&{5}(99+1)+ \\&&{2}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {964523}= &&\gray{9\cdot99999}+ \\&&\gray{6\cdot9999}+ \\&&\gray{4\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{2\cdot9}+ \\&& {9}+{6}+{4}+{5}+{2}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${964523}$ is divisible by $9$ if ${ 9}+{6}+{4}+{5}+{2}+{3}$ is divisible by $9$ Add the digits of ${964523}$ $ {9}+{6}+{4}+{5}+{2}+{3} = {29} $ If ${29}$ is divisible by $9$ , then ${964523}$ must also be divisible by $9$ ${29}$ is not divisible by $9$, therefore ${964523}$ must not be divisible by $9$.